| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola area calculations |
| Difficulty | Challenging +1.2 This is a Further Maths FP1 question requiring parametric parabola tangent equations, solving simultaneous equations to find p and q, then calculating a triangle area. While it involves multiple steps and coordinate geometry with parametric forms, the techniques are standard FP1 material with straightforward algebraic manipulation once the tangent formula is recalled. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y^2 = 16x \Rightarrow 2y\frac{dy}{dx} = 16 \Rightarrow \frac{dy}{dx} = \frac{8}{y} = \frac{8}{4p}\) OR \(y^2 = 16x \Rightarrow \frac{dy}{dx} = 2x^{-\frac{1}{2}} = 2(p^2)^{-\frac{1}{2}}\) OR \(\frac{dy}{dx} = \frac{dy}{dp}\frac{dp}{dx} = \frac{4}{2p}\) | M1 | 3.1a — Attempts differentiation to obtain \(\frac{dy}{dx}\) in terms of \(p\) or \(q\). Requires \(\alpha y\frac{dy}{dx} = \beta\) or \(\frac{dy}{dx} = \alpha x^{-\frac{1}{2}}\) or chain rule approach |
| \(\frac{dy}{dx} = \frac{8}{4p} \Rightarrow y - 4p = \frac{2}{p}(x - p^2)\) or \(y - 4q = \frac{2}{q}(x - q^2)\) | M1 A1 | 3.1a, 1.1b — M1: Correct straight line method using \(P\) or \(Q\). A1: Correct general tangent at \(P\) or \(Q\) or both |
| Using \(x = -28\) and \(y = 6\): \(6p = -56 + 2p^2 \Rightarrow p = \ldots\) | M1 | 3.1a — Uses \(x=-28\), \(y=6\) correctly placed in tangent equation and solves cubic for 2 values of \(p\) (or \(q\)) |
| \(p \text{ (or } q) = -4, 7\) | A1 | 1.1b — Correct values |
| \((16, -16),\ (49, 28)\) | A1 | 2.2a — Correct coordinates of \(P\) and \(Q\) |
| Way 1: \(\frac{1}{2}\begin{vmatrix}-28 & 16 & 49 & -28 \\ 6 & -16 & 28 & 6\end{vmatrix} = \frac{1}{2}\ | 448+448+294-96+784+784\ | \) |
| Way 2: \(77\times44 - \frac{1}{2}\times44\times22 - \frac{1}{2}\times77\times22 - \frac{1}{2}\times44\times33\) | M1 | 3.1a |
| Way 3: \(\frac{1}{2}\cdot22\sqrt{5}\cdot11\sqrt{53}\sin\!\left(\cos^{-1}\!\left(\frac{(11\sqrt{53})^2+(22\sqrt{5})^2-55^2}{2\times11\sqrt{53}\times22\sqrt{5}}\right)\right)\); NB angle at \(R\) is 42.5° | M1 | 3.1a |
| Way 8: \(\frac{1}{2}\ | RP\times QP\ | = \frac{1}{2}\left |
| \(= 1331\) (units²) | A1* | 1.1b — Correct area. Allow even if candidate reverts to decimals within solution, providing all working is correct |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y^2 = 16x \Rightarrow 2y\frac{dy}{dx} = 16 \Rightarrow \frac{dy}{dx} = \frac{8}{y} = \frac{8}{4p}$ OR $y^2 = 16x \Rightarrow \frac{dy}{dx} = 2x^{-\frac{1}{2}} = 2(p^2)^{-\frac{1}{2}}$ OR $\frac{dy}{dx} = \frac{dy}{dp}\frac{dp}{dx} = \frac{4}{2p}$ | M1 | 3.1a — Attempts differentiation to obtain $\frac{dy}{dx}$ in terms of $p$ or $q$. Requires $\alpha y\frac{dy}{dx} = \beta$ or $\frac{dy}{dx} = \alpha x^{-\frac{1}{2}}$ or chain rule approach |
| $\frac{dy}{dx} = \frac{8}{4p} \Rightarrow y - 4p = \frac{2}{p}(x - p^2)$ or $y - 4q = \frac{2}{q}(x - q^2)$ | M1 A1 | 3.1a, 1.1b — M1: Correct straight line method using $P$ or $Q$. A1: Correct general tangent at $P$ or $Q$ or both |
| Using $x = -28$ and $y = 6$: $6p = -56 + 2p^2 \Rightarrow p = \ldots$ | M1 | 3.1a — Uses $x=-28$, $y=6$ correctly placed in tangent equation and solves cubic for 2 values of $p$ (or $q$) |
| $p \text{ (or } q) = -4, 7$ | A1 | 1.1b — Correct values |
| $(16, -16),\ (49, 28)$ | A1 | 2.2a — Correct coordinates of $P$ and $Q$ |
| **Way 1:** $\frac{1}{2}\begin{vmatrix}-28 & 16 & 49 & -28 \\ 6 & -16 & 28 & 6\end{vmatrix} = \frac{1}{2}\|448+448+294-96+784+784\|$ | M1 | 3.1a — Completes a correct complete method for area of $PQR$ |
| **Way 2:** $77\times44 - \frac{1}{2}\times44\times22 - \frac{1}{2}\times77\times22 - \frac{1}{2}\times44\times33$ | M1 | 3.1a |
| **Way 3:** $\frac{1}{2}\cdot22\sqrt{5}\cdot11\sqrt{53}\sin\!\left(\cos^{-1}\!\left(\frac{(11\sqrt{53})^2+(22\sqrt{5})^2-55^2}{2\times11\sqrt{53}\times22\sqrt{5}}\right)\right)$; NB angle at $R$ is 42.5° | M1 | 3.1a |
| **Way 8:** $\frac{1}{2}\|RP\times QP\| = \frac{1}{2}\left|\binom{77}{22}\times\binom{33}{44}\right| = \frac{1}{2}(2662)$ | M1 | 3.1a — For such methods, minimum of e.g. $\frac{1}{2}(2662)$ must be seen |
| $= 1331$ (units²) | A1* | 1.1b — Correct area. Allow even if candidate reverts to decimals within solution, providing all working is correct |
---\begin{enumerate}
\item The parabola $C$ has equation
\end{enumerate}
$$y ^ { 2 } = 16 x$$
The distinct points $P \left( p ^ { 2 } , 4 p \right)$ and $Q \left( q ^ { 2 } , 4 q \right)$ lie on $C$, where $p \neq 0 , q \neq 0$\\
The tangent to $C$ at $P$ and the tangent to $C$ at $Q$ meet at the point $R ( - 28,6 )$.\\
Show that the area of triangle $P Q R$ is 1331
\hfill \mbox{\textit{Edexcel FP1 2019 Q4 [8]}}